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How do you solve the system of equations #3x + 5y - 30 = 0# and #x - y - 2 = 0#?

Answer 1

Ethan Adkins

#x = 5#
#y=3#

#3x+5y−30=0#---------------(1)
#x−y−2=0# -------------------(2)

Solve the equation (2) for #y#

#-y = 2-x#
#y = -2+x#

Substitute the value of #y = -2+x# in equation (1)
#3x+5( -2+x)−30=0#

Simplify

#3x -10+5x −30=0#
#8x −40=0#
#8x =40#
#x=40/8=5#
#x = 5#

Substitute #x = 5# in equation (2)

#5−y−2=0#

Simplify -
#3-y=0#
#-y =-3#
#y=3#

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Answer 2

Eva Abramson

To solve the system of equations:

[ 3x + 5y - 30 = 0 ] [ x - y - 2 = 0 ]

First, solve the second equation for ( x ):

[ x = y + 2 ]

Now substitute this expression for ( x ) into the first equation:

[ 3(y + 2) + 5y - 30 = 0 ]

Simplify and solve for ( y ):

[ 3y + 6 + 5y - 30 = 0 ] [ 8y - 24 = 0 ] [ 8y = 24 ] [ y = 3 ]

Now that we have the value of ( y ), substitute it back into ( x = y + 2 ):

[ x = 3 + 2 ] [ x = 5 ]

So the solution to the system of equations is ( x = 5 ) and ( y = 3 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.